Let $a(x)=x^4-8x^2+2x+3$, and $b(x)=x$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Solution: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{x^4-8x^2+2x+3}{x}&=\dfrac{ {x^4}-8 {x^2}+2 x}{ x}+\dfrac{3}{x}\\\\ &={x^3-8x+2}+\dfrac{{3}}{x}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${3}$ is less than the degree of $x$, it follows that ${r(x)}={3}$, and ${q(x)}={x^3-8x+2}$. To conclude, $q(x)=x^3-8x+2$ $r(x)=3$ [Is there another way of doing this?]